Question

the sum of the the square of two positive integer is 208 if the square of the larger number is 18 times the smaller number, find the numbers​

Answer 1

Answer:

LGBT fight b haha at Jeff tdnd

Step-by-step explanation:

with y did the same thing as too much to ask

Answer 2

Answer:

x = 12 and y = 8

Step-by-step explanation:

Let the two positive integers be x and y, where x > y.

Sum of the squares of x and y is 208.

$x^2 + y ^2 = 208$     $\rightarrow Eqn. (1)$

Square of x is 18 times y.

$x^2 = 18y$             $\rightarrow Eqn. (2)$

Substituting the value of $x^2$ into (1):

$18y+y^2=208$

$y^2 + 18y - 208 = 0$

$[ 26 \times 8 = 208;$  $26 - 8 = 18 ]$

$y^2 + 26y - 8y -208 = 0$

$y(y+26) - 8(y+26) = 0$

$(y+26)(y-8)=0$

$y = \{-26, +8\}$

But y is a positive integer. Therefore $y =+8$

Substituting the value of $y$ into (2):

$x^2 = 18(8)$

$x^2 = 144$

$x=\sqrt{144}$

$x=\{-12, +12\}$

But x is a positive integer. Therefore $x =+12$

Any corrections or suggestion for improvement are welcome :)